# Math Help - Trig Function

1. ## Trig Function

$C(t) = 38+2sin(\frac{\pi}{6}t)$ models a patient in degrees celsius during the 12 days of illness.

It asks to use the function C(t) to find the maximum temperature that the person reaches during the illness.

$C(t) = 38+2sin(\frac{\pi}{6}t)$

Differentiated and equate to zero

$C'(t)= 2cos\Big(\frac{\pi}{6}t\Big)*\frac{\pi}{6}$

$0 = 2cos\Big(\frac{\pi}{6}t\Big)*\frac{\pi}{6}$

$t = \Big(\frac{cos^{-1}(0)}{\frac{\pi}{6}}\Big)$

$t = 3$
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$C(t) = 38+2sin(\frac{\pi}{6}3)$

$C(t) = 40$

My question is, I've found the maximum for the modelled function (3,40), now what if they asked to also find the minimum. I know the amplitude is 2 therefore the minimum will be 36 degrees celsius, (9, 36). However, that is just by observation of the features of the function. How do you find it algebraically?

2. Originally Posted by dwat
$C(t) = 38+2sin(\frac{\pi}{6}t)$ models a patient in degrees celsius during the 12 days of illness.

It asks to use the function C(t) to find the maximum temperature that the person reaches during the illness.

$C(t) = 38+2sin(\frac{\pi}{6}t)$

Differentiated and equate to zero

$C'(t)= 2cos\Big(\frac{\pi}{6}t\Big)*\frac{\pi}{6}$

$0 = 2cos\Big(\frac{\pi}{6}t\Big)*\frac{\pi}{6}$

$t = \Big(\frac{cos^{-1}(0)}{\frac{\pi}{6}}\Big)$

$t = 3$
-----------------------------------------------------------------
$C(t) = 38+2sin(\frac{\pi}{6}3)$

$C(t) = 40$

My question is, I've found the maximum for the modelled function (3,40), now what if they asked to also find the minimum. I know the amplitude is 2 therefore the minimum will be 36 degrees celsius, (9, 36). However, that is just by observation of the features of the function. How do you find it algebraically?
there is infinity many solution for
$t = \Big(\frac{cos^{-1}(0)}{\frac{\pi}{6}}\Big)$

$t = 3$

you can take

$cos(\frac{\pi}{6}t)=0 ..so.....\frac{\pi}{6}t=\frac{\pi}{2}..or...\frac{ \pi}{6}t=\frac{3\pi}{2}...$

so $t=\frac{6(3\pi)}{2\pi}$

so t=9 right ...